28.11 a) y = 1/x + 4x; б) y = -2√x - 1/x; в) y = 1/x - 6x; г) y = 8√x + 1/x.

Решение:

28.11

в) $$y = \frac{1}{x} - 6x$$

$$y' = (\frac{1}{x})' - (6x)' = -\frac{1}{x^2} - 6$$

г) $$y = 8\sqrt{x} + \frac{1}{x}$$

$$y' = (8\sqrt{x})' + (\frac{1}{x})' = 8 \cdot \frac{1}{2\sqrt{x}} - \frac{1}{x^2} = \frac{4}{\sqrt{x}} - \frac{1}{x^2}$$

Ответ: в) $$y' = -\frac{1}{x^2} - 6$$, г) $$y' = \frac{4}{\sqrt{x}} - \frac{1}{x^2}$$